The DAV Maths Book Class 4 Solutions and DAV Class 4 Maths Chapter 11 Worksheet 3 Solutions of Perimeter offer comprehensive answers to textbook questions.
DAV Class 4 Maths Ch 11 WS 3 Solutions
Question 1.
Solve the following word problems.
(а) Find the perimeter of a rectangle whose length is 12 cm and breadth is 8 cm.
Answer:
Perimeter of a rectangle = 2 l + 2 b
= 2 × 12 cm + 2 × 8 cm
= 24 cm + 16 cm
= 40 cm
Perimeter = 40 cm
(b) Calculate the perimeter of the square whose sides are given below.
(i) 6 cm
(ii) 11 meters
(iii) 26 cm
Answer:
(i) Perimeter of square = 4 × side = 4 × 6 cm
= 24 cm.
(ii) Perimeter of square = 4 × side = 4 × 11 m
= 44 m.
(iii) Perimeter of square = 4 × side = 4 × 26 cm
= 104 cm.
(c) The length of three sides of a triangle are 6 cm, 8 cm and 10 cm. Find its perimeter.
Answer:
Perimeter of triangle = sum of three sides
= 6 cm + 8 cm + 10 cm
= 24 cm.
(d) A boy runs around a rectangular field, the length of which is 35 metres and the breadth is 20 metres. How much distance will be run?
Answer:
To find the distance, we have to find perimeter of rectangular field.
Length of rectangular field = 35 m
Breadth of rectangular field = 20 m
Perimeter of field = 2 × 35m + 2 × 20m
= 70 m + 40 m = 110 m
Boy will run 110 m.
(e) Sheela wants to fix ribbon along the border of a painting. If the length and breadth of the painting are 36 cm and 15 cm respectively how much ribbon is required by Sheela?
Answer:
Since the length and breadth is different so shape of frame is rectangle so we have to find perimeter of frame.
Length of frame = 36 cm
Breadth of frame = 15 cm
Perimeter of frame = 2 × 36cm + 2 × 15cm
= 72 cm + 30 cm
= 102 cm
102 cm ribbon is required.
(f) Raju has to frame a square photo having one side of length 12 cm. How much material will Raju require for this job?
Answer:
We have to find the perimeter of square photo.
Perimeter of square frame = 4 × side
= 4 × 12 cm
= 48 cm
48 cm material is required.
(g) A boy runs two times around a field, the length of which is 62 m and breadth is 40 m. Find the distance covered by him.
Answer:
We have to find perimeter of field.
Length of field = 62 m
Breadth of field = 40 m
Perimeter of the field = 2(62 + 40)
= 2(102)
= 204 m
He runs 2 times so distance covered = 2 × 204 m
= 408 m.
(h) Tina ran around a rectangular garden of length 12 m and breadth 8 m. Meena ran around a square field of side 11m. Who covered more distance and by how much?
Answer:
In both cases we have to find distance means perimeter.
Distance covered by Tina around a rectangular garden = 2 × 12 m + 2 × 8 m
= 24 m + 16 m
= 40 m
Distance covered by Meena, around a square field = 4 × side
= 4 × 11 m
= 44 m
Hence, Meena covered more distance.
Extra distance = 44 – 40
= 4 m
DAV Class 4 Maths Chapter 11 Worksheet 3 Notes
Perimeter of Rectangle, Square And Triangle
(i) Rectangle:
Opposite sides of the rectangle are equal so the perimeter will be
= l + b + l + b
= 2l + 2 b
= 2 (l + b) unit
Let l = 3 cm, b = 2 cm,
Perimeter 2(3 + 2)
= 10 cm
(ii) Square:
Let the side of square be a since all the sides are equal.
The perimeter = a + a + a + a.
= 4a unit
Let side = 4 cm
Perimeter = 4 × 4
= 16 cm
(iii) Triangle
Perimeter of the triangle will be sum of three sides.
Let the sides be a, b, and c
Perimeter = a + b + c
Let a = 2cm, b = 3cm and c = 4cm
Perimeter = 2 + 3 + 4
= 9 cm.
Question 1.
Babita’s grandfather has a rectangular garden whose length is 70 m and breadth is 40 metres. How much wire is needed to fence the field?
Answer:
The length of wire needed will be equal to perimeter of field.
Length of field = 70 m
Breadth of field = 40 m
Length of wire needed = 2 × 70m + 2 × 40m
= 140 m + 80 m
= 220 m
220 m wire is needed.
Question 2.
Meena wants to fix a border of a square carpet whose each side is 9 m. Find the length of border.
Answer:
Length of border is equal to perimeter of square.
Each side of carpet = 9 m
Length of border = 4 × 9 m = 36 m
36 m length is needed to border the carpet.